6:54 PM Reader View Available Verizon LTE 11. Consider the following questions a
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6:54 PM Reader View Available Verizon LTE 11. Consider the following questions about CIs. A researcher tests emotional intelligence (EI) for a random sample of children selected from a population of all students who are enrolled in a school for gifted children. The researcher wants to estimate the mean EI for the entire school. The population standard deviation, o, for El is not known. Let's suppose that a researcher wants to set up a 95% CI for IQ scores using the following information The sample mean M 130 The sample standard deviation s = 15 The sample size N= 120 The d-N-1 = 119 For the values given above, the limits of the 95% CI are as followS: Lower limit = 130-1.96 × 1.37-127.31; Upper limit = 130 + 1.96 × 1.37 = 132.69 The following exercises ask you to experiment to see how changing some of the values involved in computing the CI influences the width of the CI Recalculate the CI above to see how the lower and upper limits (and the width of the CI) change as you vary the N in the sample (and leave all the other values the same. a. What are the upper and lower limits of the CI and the width of the 95% CI if all the other values remain the same (M= 130, s-15) but you changeExplanation / Answer
a) M= 130, s= 15 and N=16
Calculation M = 130 t = 2.13 and sM = (152/16) = 3.75
= M ± t(sM)
= 130 ± 2.13*3.75
= 130 ± 7.99
M = 130, 95% CI [122.01, 137.99].
You can be 95% confident that the population mean () falls between 122.01 and 137.99.
b) N=25
M = 130
t = 2.06
sM = (152/25) = 3
= M ± t(sM)
= 130 ± 2.06*3
= 130 ± 6.19
M = 130, 95% CI [123.81, 136.19].
You can be 95% confident that the population mean () falls between 123.81 and 136.19.
c) N=49 M = 130 t = 2.01 and sM = (152/49) = 2.14
= M ± t(sM)
= 130 ± 2.01*2.14
= 130 ± 4.31
M = 130, 95% CI [125.69, 134.31].
You can be 95% confident that the population mean () falls between 125.69 and 134.31.
d) As N=16 width of 95% ci is 15.98
N=25 width of 95% c.i is 12.38
N=49 width of 95% c.i is 8.62
Here as N increases width of confidence interval decreases.
e) CI =80% M=130 s= 15 and N=49
Calculation M = 130 t = 1.3 and sM = (152/49) = 2.14
= M ± t(sM)
= 130 ± 1.3*2.14
= 130 ± 2.78
M = 130, 80% CI [127.22, 132.78].
You can be 80% confident that the population mean () falls between 127.22 and 132.78.
width of 80% c.i is [132.78-127.22] =5.56
f) C.I = 99% , M= 130 s=15 and N=49
t = 2.68
sM = (152/49) = 2.14
= M ± t(sM)
= 130 ± 2.68*2.14
= 130 ± 5.75
M = 130, 99% CI [124.25, 135.75].
You can be 99% confident that the population mean () falls between 124.25 and 135.75.
Width of 99% c.i is [135.75-124.25]= 11.5
g) On increasing the level of the confidence the width of confidence interval increases.
Please Note : I have done all the parts from a to g . Please post the question 12 again with data. Please comment for any doubt. Thank you!
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